Homework 37 - portillo (gdp347) homework 37 Turner (58185)...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: portillo (gdp347) homework 37 Turner (58185) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 4) 10.0 points A traveling wave propagates according to the expression y = (4 . 39 cm) sin bracketleftBig (1 . 85 cm 1 ) x- (3 . 5 s 1 ) t bracketrightBig , where x is in centimeters, and t is in seconds. Find the amplitude of the wave. Correct answer: 4 . 39 cm. Explanation: Let : A = 4 . 39 cm . Given a wave y = A sin( k x- t ), the ampli- tude of the wave is A = 4 . 39 cm . 002 (part 2 of 4) 10.0 points Determine the wavelength of the wave. Correct answer: 3 . 39632 cm. Explanation: Let : k = 1 . 85 cm 1 . The angular wave number of the wave is k , so = 2 k = 2 1 . 85 cm 1 = 3 . 39632 cm . 003 (part 3 of 4) 10.0 points Determine the frequency of the wave. Correct answer: 0 . 557042 Hz. Explanation: Let : = 3 . 5 s 1 . The angular frequency of the wave is , so f = 2 = 3 . 5 s 1 2 = . 557042 Hz . 004 (part 4 of 4) 10.0 points Determine the period of the wave. Correct answer: 1 . 7952 s. Explanation: The period of the wave is T = 1 f = 1 . 557042 Hz = 1 . 7952 s . 005 10.0 points The red light emitted by a He-Ne laser has a wavelength of 633 nm in air and travels at 3 . 00 10 8 m/s. Find the frequency of the laser light. Correct answer: 4 . 73934 10 14 Hz. Explanation: Let : = 633 nm and v = 3 10 8 m / s . The speed of light is given by v = f f = v = 3 10 8 m / s 633 nm 10 9 nm 1 m = 4 . 73934 10 14 Hz . 006 10.0 points A wave on a string is described by the wave function y = (0 . 1 m) sin[(0 . 5 rad / m) x- (20 rad / s) t ] Determine the frequency of oscillation of a particular point at x = 2 . 0 m. Correct answer: 3 . 1831 Hz. Explanation: Let : = 20 rad / s . In fact, when a wave with frequency f trav- els along a string, any point of the string has the same oscillation frequency f . In this case, f = 2 = 20 rad / s 2 = 3 . 1831 Hz keywords: 007 (part 1 of 2) 10.0 points A steel piano wire is 0 . 7 m long and has a...
View Full Document

This note was uploaded on 01/25/2012 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

Page1 / 5

Homework 37 - portillo (gdp347) homework 37 Turner (58185)...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online